Problem 1 :
It is a graph in which each value is shown as a __________ (or and x) above a number line. Each dot (or x) represents __________ response.
1) How many people read 4 books last summer?
2) How many people read 1 book last summer?
3) How many people were surveyed?
4) Only 2 people read _____ books last summer
Solution :
It is a graph in which each value is shown as a number of books above a number line.
Each dot (or x) represents the number of people response.
1) 2 people read 4 books last summer.
2) Above 1, we have 6 dots. So, 6 people read 1 book last summer.
3) Total number of people surveyed :
= 0(4)+1(6)+2(3)+3(1)+4(2)+5(1)+6(0)+7(1)+8(1)+9(1)
= 0+6+6+3+8+5+0+7+8+9
= 52
So, 52 people were surveyed.
4) 2 people read 4 books.
Problem 2 :
Use the data in the dot plot to answer questions.
1) What is the mean number of shells collected?
2) What is the median number of shells collected?
3) What is the mode?
4) What is the range?
Solution :
1)
So, the mean is 3.
2) Median :
Number of data values (N) = 13 (odd)
= (N + 1)/2 th value
= 7th value
So, 3 in the median.
3) Mode :
Most repeated value is 1. So, mode is 1.
4) Range :
Largest value = 5, smallest value = 1
Range = 5 - 1 ==> 4
Problem 3 :
Fourteen students were surveyed about the time they spend exercising and playing video games each week.
Compare the data by answering the questions.
1) What is the range for the hours of exercise?
For playing video games?
2) What is the mode for exercise? Playing video games?
3) What is the median hours spent exercising? Playing video games?
4) What is the mean number of hours spent exercising? Playing video games?
Solution :
1) For exercise (h) :
Largest value = 12, smallest value = 0
Range = 12 - 0 ==> 12
For video games (h) :
Largest value = 12, smallest value = 0
Range = 14 - 0 ==> 14
2) Exercise (h) :
1 and 2 are repeated 3 times. So. mode is 1 and 2.
Video games :
6, 8, 9, 11 and 12 are repeated 2 times. So, mode are 6, 8, 9, 11 and 12.
3) Median :
Total number of values = 14 (even)
So, median will be, average of 7th value and 8th value.
Exercise (h) = (2 + 3)/2 = 2.5 hours |
Video games (h) = (9 + 9)/2 = 9 hours |
4) Mean :
Exercise :
= [0(1)+1(3)+2(3)+3(3)+5(1)+7(2)+12(2)] / 14
= 61 / 14
= 4.35
Video games :
= [0(1) + 6(2)+7(1)+8(2)+9(2)+10(1)+11(2)+12(2)+14(1)]/14
= 123/14
= 8.78 hours
Problem 4 :
Mrs. Andrews surveyed the students in her math classes to fnd out how many minutes each one studied for the last test. She created a dot plot for 1st period and for 2nd period.
Which of the following statements are supported by the data in the two dot plots?
I. The least number of minutes students studied in Mrs. Andrews’ 1st period is greater than the least number of minutes students studied in Mrs. Andrews’ 2nd period.
II. The data for Mrs. Andrews’ 2nd period is more symmetrical than the data for Mrs. Andrews’ 1st period.
III. The mode number of minutes students studied in Mrs. Andrews’ 1st period is less than the mode number of minutes students studied in Mrs. Andrews’ 2nd period.
IV. The range of the data for Mrs. Andrews’ 1st period is less than the range of the data for Mrs. Andrews’ 2nd period.
V. The median number of the data for Mrs. Andrews’ 2nd period is greater than the median number of the data for Mrs. Andrews’ 1st period.
VI. The data for Mrs. Andrews’ 1st period is skewed more to the right than the data for Mrs. Andrews’ 2nd period
Solution :
Statement I :
The least number of minutes studied in Mrs. Andrews’ 1st period = 0
The least number of minutes studied in Mrs. Andrews’ 2nd period = 0
0 > 0 (false)
So, statement I is false.
Statement II :
From the above picture, it is clear statement II is true.
Statement III :
Mode of 1st period :
10 and 15 are mode of 1st period.
Mode of 2nd period :
30 is the mode of 2nd period.
Conclusion :
Mode of the 1st period is lesser than mode of 2nd period. So, statement III is true.
Statement IV :
Range of 1st period :
60 - 0 ==> 60
Range of 2nd period :
60 - 0 ==> 60
60 is not less than 60, they are equal. So, statement IV it is false.
Statement V :
Median of 1st period :
Total number of dots = 32 (even)
Average of 16th and 17th value is median.
= (11+11)/2
= 11
Median of 2nd period :
= (30+30)/2
= 30
Median of 2nd period is greater than median of 1st period. So, statement IV is true.
Statement V :
Graph of of 1st period :
Skewed right.
Graph of of 2nd period :
Symmetric
So, statement VI is true.
Problem 5 :
The double dot plot shows the numbers of junk emails that were received by Marcus and Anders for the past twenty days. Which inference about the two populations is true?
A) Both sets of data have the same interquartile range.
B) Both sets of data have the same median.
C) Anders’s data centers around 6 and Marcus’s data centers around 7.
D) The interquartile range for Anders’s data is 0.5 greater than the interquartile range for Marcus’s data.
Solution :
Interquartile range of Ander :
1, 2, 2, 3, 3, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 7
Total number of values = 20
Median is average of 10th and 11th value.
IQR = Q3 - Q1
1, 2, 2, 3, 3, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 7
Q2 = 5
1, 2, 2, 3, 3, 5, 5, 5, 5, 5 = (3+5)/2 Q1 = 4 |
5, 6, 6, 6, 6, 7, 7, 7, 7, 7 = (6+7)/2 Q3 = 6.5 |
Q3 - Q1 = 6.5 - 4
IQR = 2.5
Interquartile range of Marcus’s :
2, 3, 5, 6, 6, 6, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 9, 10, 10
Q2 = 7
2, 3, 5, 6, 6, 6, 7, 7, 7, 7 = (6 + 6)/2 Q1 = 6 |
7, 7, 8, 8, 8, 8, 8, 9, 10, 10 = (8 + 8)/2 Q3 = 8 |
Q3 - Q1 = 8 - 6
IQR = 2
D) The interquartile range for Anders’s data is 0.5 greater than the interquartile range for Marcus’s data is correct.
Problem 6 :
The dot plots show the number of miles run per week for two different classes.
Based on the data in the dot plots, write the following:
Solution :
Range :
Class A Range = 14 - 2 = 12 |
Class B Range = 9 - 4 = 5 |
Range of class A is greater than range of class B.
Median :
In both class A and B, we have 15 data values.
15 (odd)
= (15 + 1)/2 th value
= 8th value is median
Median of class A = 3
Median of class B = 7
Median of class A is 4 lesser than class B.
Mode :
Most repeated value in class A = 2
Most repeated value in class B = 7
Mode of class B is lesser than class A.
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM